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The problem in the image describes a 10 ft long ladder resting against a vertical wall. The bottom of the ladder is sliding away from the wall at a rate of \(1.1 \, \text{ft/s}\), and we are asked to determine how fast the angle between the ladder and the ground is changing (in radians per second) when the bottom of the ladder is 6 ft from the wall.

This is a related rates problem in calculus, and the relationship between the sides of the right triangle formed by the ladder, the wall, and the ground can be analyzed using trigonometry and the Pythagorean theorem.

### Step-by-step solution:

1. **Variables**:
   - Let \( x \) be the distance between the bottom of the ladder and the wall (on the ground).
   - Let \( y \) be the distance between the top of the ladder and the ground (height along the wall).
   - Let \( \theta \) be the angle between the ladder and the ground.
   - Given: the length of the ladder is 10 ft, so by the Pythagorean theorem:
     \[
     x^2 + y^2 = 10^2 \quad \Rightarrow \quad x^2 + y^2 = 100
     \]
   - The rate at which the bottom of the ladder is moving away from the wall is \( \frac{dx}{dt} = 1.1 \, \text{ft/s} \).
   - We are asked to find the rate of change of the angle, i.e., \( \frac{d\theta}{dt} \), when \( x = 6 \, \text{ft} \).

2. **Relating the variables**:
   The angle \( \theta \) can be related to \( x \) using the trigonometric function:
   \[
   \cos(\theta) = \frac{x}{10}
   \]
   Differentiating both sides with respect to time \( t \) gives:
   \[
   -\sin(\theta) \frac{d\theta}{dt} = \frac{1}{10} \frac{dx}{dt}
   \]
   Thus, solving for \( \frac{d\theta}{dt} \):
   \[
   \frac{d\theta}{dt} = -\frac{\frac{dx}{dt}}{10 \sin(\theta)}
   \]

3. **Finding values for \( \theta \) and \( \sin(\theta) \)**:
   At the instant when \( x = 6 \, \text{ft} \), we can use the Pythagorean theorem to find \( y \):
   \[
   6^2 + y^2 = 100 \quad \Rightarrow \quad y^2 = 64 \quad \Rightarrow \quad y = 8 \, \text{ft}
   \]
   Now, using trigonometry, \( \sin(\theta) \) is given by:
   \[
   \sin(\theta) = \frac{y}{10} = \frac{8}{10} = 0.8
   \]

4. **Substitute known values**:
   Now, substituting \( \frac{dx}{dt} = 1.1 \, \text{ft/s} \) and \( \sin(\theta) = 0.8 \) into the equation for \( \frac{d\theta}{dt} \):
   \[
   \frac{d\theta}{dt} = -\frac{1.1}{10 \times 0.8} = -\frac{1.1}{8} = -0.1375 \, \text{rad/s}
   \]

### Final Answer:
The rate at which the angle between the ladder and the ground is changing is approximately \(-0.1375 \, \text{rad/s}\). The negative sign indicates that the angle is decreasing.

Would you like any further explanation or details on any of the steps? 

Here are 5 related questions you might explore:
1. How would the rate of change of the angle differ if the ladder were longer?
2. How can we compute the rate of change of the height of the ladder against the wall?
3. What happens to the rate of change of the angle as the ladder moves further from the wall?
4. How would the calculation change if the rate of sliding away from the wall were variable?
5. Can you derive an expression for the rate of change of the height of the ladder at any given moment?

**Tip:** When working with related rates problems, always remember to carefully identify the variables and their relationships before differentiating implicitly.

A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.1 ft/s, how fast (in rad/s) is the angle (in radians) between the ladder and the ground changing when the bottom of the ladder is 6 ft from the wall?

Learn how to calculate the rate of change of the angle between a ladder and the ground as it slides away from a wall. This related rates problem covers the Pythagorean theorem, implicit differentiation, and trigonometric functions. Suitable for calculus students.

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